I have this function:
$$f(A,B,C,D) = A'B' + CD' + ABC + A'B'CD' + ABCD'$$
I used a Karnaugh map to minimize the function to:
$$Minimum SOP = A'B' + C D' + A B C$$
How can I turn this into a product-of-sums?
My attempt was to first find the inverse:
$$(A'B' + C D' + A B C)'$$
$$= (A'B')'(C D')'(A B C)'$$
$$= (A + B)(C' + D)(A' + B' + C')$$
$$= (AC' + AD + BC' + CD)(A' + B' + C')$$
$$= A'(AC' + AD + BC') + B'(AC' + AD + BC') + C'(AC' + AD + BC')$$
I am stuck here because I see that I end up with this:
$$= A'AC' + A'AD + A'BC' + …$$
How can this be correct?
Doesn't $A'A$ mean $A \wedge \neg A$ (which doesn't even make sense to me)?
If I made a mistake, where did I make it?
If I'm on the right track, what is my next step?
Best Answer
$$\begin{array}{r|c|c|c|c} CD|AB&00&01&11&10 \\\hline 00&1&0&0&0\\\hline 01&1&0&0&0\\\hline 11&1&0&1&0\\\hline 10&1&1&1&1 \end{array}$$
From this K-map, those $0$'s (the negative of $f$) can be written as a sum of product:
$$[f(A, B, C, D)]' = BC' + AC' + A'BD + AB'D$$
Then, using de Morgan's laws, the negative of the whole formula is
$$\begin{align*} f(A,B,C,D) =& (BC' + AC' + A'BD + AB'D)'\\ =&(BC')'(AC')'(A'BD)'(AB'D)'\\ =&(B'+C)(A'+C)(A+B'+D')(A'+B+D') \end{align*}$$
In short, you should focus on the $0$'s when finding a product of sum form.